Answer to Question #204227 in Mechanics | Relativity for harry

Question

cold storage compartment is 4 m diameter and 4.5 m long with a thickness of 75 mm with insulating material which has a coefficient of thermal conductivity of 5.8 x10^-3 W/m-K. Calculate the quantity of heat energy leaking through the insulation per hour when the outside and inside temperature of the insulation are 15 R and -5 C.

Answer 1

Gives

Thickness(x)=75mm

Length (h)=4.5m

Radius (r)=2m

Thermal conductivity $K=5.8\times10^{-3}W/m-K$

Area(A)=$\pi r^2h$

$A=3.14\times4\times4.5=56.52m^2$

Out side temperature (T_o)=25R=18.75°C

Inside temperature (T_i)=-5°C

Now we can written as

Rate of heat

$$\frac{Q}{t}=\frac{KA(T_o-T_i)}{x}$$

Put value

$\frac{Q}{t}=\frac{5.8\times10^{-3}\times{56.52}(18.75-(-5))}{75\times10^{-3}}$

$\frac{Q}{t}=103.809$ J/sec

Heat leakage per hour

$Q=\frac{103.809}{3600}=0.0289$ J/hour